Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 2 14D d Solution Created 2026-09-24 Updated 2026-09-25
Permuting the gaps, or equivalently relabelling the particles, leaves both quadratic forms invariant. The three-dimensional permutation representation splits as the rigid-rotation line spanned by and the two-dimensional sum-zero standard representation of the symmetric group.
For three unit masses, andThe uniform vector has , whereas every vector with coordinate sum zero is annihilated by . One orthonormal choice of normal modes is thereforeIf is retained from (b), the angular frequencies are . The equal upper frequencies follow from degenerate normal modes from permutation symmetry: the stiffness is scalar on the two-dimensional sum-zero subspace. In the uniform mode the inter-bar strings do not change length, but all three beam connections stretch, giving the nonzero lower frequency. The motion is an arbitrary sine-cosine superposition of these modes about the gravitational equilibrium.
A common shift of all three angles leaves every spring length unchanged. This continuous rotational symmetry gives the zero mode: there is no restoring force against rigid rotation.
The equilateral equilibrium also has the dihedral group symmetry . Its action on the three angular displacements preserves the uniform one-dimensional subspace and the two-dimensional sum-zero subspace. The latter is an irreducible representation over the real numbers; a reflection may reverse all angular signs as well as exchange bead labels, which does not change these invariant subspaces. The symmetric stiffness matrix must act as a scalar on the sum-zero subspace, explaining the equal vibrational angular frequencies. Directly, each sum-zero vector satisfies . This is an example of degenerate normal modes from permutation symmetry, and of normal modes of three equal masses with a symmetric gap potential.